CodeForces 551E GukiZ and GukiZiana

GukiZ and GukiZiana

Time Limit: 10000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 551E
64-bit integer IO format: %I64d      Java class name: (Any)

 

Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and numbery, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:

1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
2. Second type has form 2 y and asks you to find value of GukiZiana(a, y).

For each query of type 2, print the answer and make GukiZ happy!

 

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries.

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a.

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

 

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

 

Sample Input

Input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3

Output

2

Input

2 3
1 2
1 2 2 1
2 3
2 4

Output

0
-1

Source

Codeforces Round #307 (Div. 2)

 

解题：分块搞

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1010;
LL a[maxn*maxn],lazy[maxn],x;
vector<int>block[maxn];
int b_size,N,pos[maxn*maxn],n,q,cmd,L,R;
bool cmp(const int x,const int y) {
    if(a[x] == a[y]) return x < y;
    return a[x] < a[y];
}
void update(int L,int R,LL x) {
    int k = pos[L],t = pos[R];
    if(k == t) {
        for(int i = L; i <= R; ++i) a[i] += x;
        sort(block[k].begin(),block[k].end(),cmp);
        return;
    }
    for(int i = k + (pos[L-1] == k); i <= t - (pos[R + 1] == t); ++i) lazy[i] += x;
    if(pos[L-1] == k) {
        for(int i = L; pos[i] == k; ++i) a[i] += x;
        sort(block[k].begin(),block[k].end(),cmp);
    }
    if(pos[R+1] == t) {
        for(int i = R; pos[i] == t; --i) a[i] += x;
        sort(block[t].begin(),block[t].end(),cmp);
    }
}
LL query(LL x) {
    int L = -1,R = -1,i;
    for(i = 1; i <= N; ++i){
        a[0] = x - lazy[i];
        vector<int>::iterator it = lower_bound(block[i].begin(),block[i].end(),0,cmp);
        if(it == block[i].end()) continue;
        if(a[*it] + lazy[i] == x){
            L = *it;
            break;
        }
    }
    if(L == -1) return -1;
    for(int j = N; j >= i; --j){
        a[n+1] = x - lazy[j];
        vector<int>::iterator it = lower_bound(block[j].begin(),block[j].end(),n+1,cmp);
        if(it == block[j].begin()) continue;
        --it;
        if(a[*it] + lazy[j] == x){
            R = *it;
            break;
        }
    }
    return R - L;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>q;
    b_size = ceil(sqrt(n*1.0));
    for(int i = 1; i <= n; ++i) {
        cin>>a[i];
        pos[i] = (i - 1)/b_size + 1;
        block[pos[i]].push_back(i);
    }
    N = (n - 1)/b_size + 1;
    for(int i = 1; i <= N; ++i) sort(block[i].begin(),block[i].end(),cmp);
    while(q--) {
        cin>>cmd;
        if(cmd == 1) {
            cin>>L>>R>>x;
            update(L,R,x);
        } else {
            cin>>x;
            cout<<query(x)<<endl;
        }
    }
    return 0;
}